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In Q1, |x - 2y| = 4 represents the two lines x - 2y - 4 = 0 and x - 2y + 4 = 0. So the inequality represents the region between these two parallel lines (test the point (0,0) to check).

$\noindent \textbf{Q2.} The roots are $\alpha$ and $2\alpha$. So $3\alpha = \frac{-b}{a} $\noindent \textbf{Q3.} Let $u = \sin{x}$ $\noindent \textbf{Q4.} $\alpha + \beta = 0$ $\noindent \textbf{Q5.} $|x| = \begin{cases} x \text{ if } x \geq 0 \\ -x \text{ if } x < 0 \end{cases} $\noindent...

The conductor (eg. copper wire)

$\noindent \textbf{(a)} Since $\angle GKC = 45\degree$ then $\triangle GKC$ is an isosceles right-angled triangle, therefore, $CK = GC = h$. In $\triangle BCK$, $\angle BCK = 180\degree - 45\degree = 135 \degree$. Now apply the cosine rule, \\ \begin{align*} \quad BK^2 &= CK^2 + BC^2 -...

$\noindent My interpretation of the question meant that the diagonal itself was 9 cm rather than the side, and the length of the other diagonal was being asked for. \\\\ If the question meant the rhombus had \textbf{side 9 cm} then it simply becomes, \\ \begin{align*}\quad \cos{16\degree} &=...

$\noindent Use the property that the diagonals of a rhombus bisect each other at right angles. Also, the diagonals of a rhombus divide the rhombus into four congruent triangles. \\\\ In any one of these triangles, if $l$ is the length of the other diagonal, \\ \begin{align*} \quad...

In (a) the construction of point P is not entirely necessary as long as you can prove angle EAD = angle BEC (which can be done multiple ways, such as alternate angles, angle sum of a triangle, angles on a straight line, etc)

$\noindent \textbf{(a)} Produce $BA$ to $P$. (if the rectangle is named beginning from the top-left corner and going clockwise) \\ \begin{align*} \angle AEC &= \angle EAP \text{ (alternate angles)} \\ \text{But }\angle AEC &= 90 + \angle BEC \\ \text{and } \angle EAP &= 90 + \angle EAD \\...

Although perhaps there is a much more elegant method. eg. (alternatively) \alpha and \beta can be used to find the lengths of the legs of the triangle made by two adjacent sides of the rectangle and the row adjacent to the diagonal. In which Pythagoras' Theorem can then be applied to find the...

$\noindent The given diagonal splits the rectangle into two congruent triangles. Both triangles are right-angled, but are not isosceles. The two angles formed by the hypotenuses of these triangles and their legs can be found using trigonometry, \\ \begin{align*} \quad \alpha &=...

360° is correct just make sure the inequality 0°≤α≤360° was not strict.

Yes I understand that no reference to units are made, hence why I said to nevermind what I said. 'Angle' is a ratio of the length of a circular arc to a radius. Both quantities involved in this ratio share the same dimensionality and so are cancelled out, naturally leaving 'angles' as...

Sorry, nevermind what I said. I guess a geometrical issue arises when you try to graph a linear function on top of a trigonometric function that has been graphed using degrees as its input. This is because an input in degrees into a trigonometric function produces an output that is not in...

$\noindent If the question is in radians, use radians. If in degrees, use degrees. $\pi$ is used instead of 180 degrees because $\pi$ radians is equivalent to 180 degrees. \\\\ The reason why we use the radian angle measure is because of many things. Most importantly, it simplifies Calculus...

$\noindent It can. If $y = x^5$ then $\frac{d^3y}{dx^3} = 60x^2$. If $\frac{d^3y}{dx^3} = 0$ then $x = 0$. And at $x = 0$ there is a point of inflexion on $y = x^5$.

No

$\noindent The green shaded area is below the curve $y = \cos{x}$ and is not bounded by the curve $y = \sin{x}$. Instead, the area is bounded by the line $x = \frac{\pi}{4}$, the curve $y = \cos{x}$ and the $x$-axis.

$\noindent $A = \int_0^{\frac{\pi}{4}}\sin{x}\mathrm{d}x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos{x}\mathrm{d}x = 2\int_0^{\frac{\pi}{4}}\sin{x}\mathrm{d}x$ (by symmetry). And so $A = (2 - \sqrt{2})$ units$^2$

$\noindent \begin{align*}\frac{f(x+h) - f(x)}{h} &= \frac{[5(x+h) + 1] - (5x+1)}{h} \\ &= \frac{5x + 5h + 1 - 5x - 1}{h} \\ &= \frac{5h}{h} \\ &= 5 \end{align*} \\ And so $f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0}5 = 5$ \\ ie. $f'(x) = 5.

The root can be undone by squaring both sides, which is the answer to your question.